Solve the following system

Answer:{x = -4 , y = 2 ,  z = 1Step-by-step explanation:Solve the following system: {-2 x + y + 2 z = 12 | (equation 1) 2 x - 4 y + z = -15 | (equation 2) y + 4 z = 6 | (equation 3) Add equation 1 to equation 2: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x - 3 y + 3 z = -3 | (equation 2) 0 x+y + 4 z = 6 | (equation 3) Divide equation 2 by 3: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x - y + z = -1 | (equation 2) 0 x+y + 4 z = 6 | (equation 3) Add equation 2 to equation 3: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x - y + z = -1 | (equation 2) 0 x+0 y+5 z = 5 | (equation 3) Divide equation 3 by 5: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x - y + z = -1 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Subtract equation 3 from equation 2: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x - y+0 z = -2 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Multiply equation 2 by -1: {-(2 x) + y + 2 z = 12 | (equation 1) 0 x+y+0 z = 2 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Subtract equation 2 from equation 1: {-(2 x) + 0 y+2 z = 10 | (equation 1) 0 x+y+0 z = 2 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Subtract 2 × (equation 3) from equation 1: {-(2 x)+0 y+0 z = 8 | (equation 1) 0 x+y+0 z = 2 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Divide equation 1 by -2: {x+0 y+0 z = -4 | (equation 1) 0 x+y+0 z = 2 | (equation 2) 0 x+0 y+z = 1 | (equation 3) Collect results: Answer:  {x = -4 , y = 2 ,  z = 1